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For two sets A,B with $B\subset A$ let us define:

$$A-B:=A\setminus B $$

as the monotone difference. If $B$ is not a subset of $A$ then $A-B$ is not defined. Now my question. How to portray $A \cup B$ with only $-$ and $\cap$?

Thanks for hints and/or solutions.

My idea: $$A \cup B=(X-A-B)^c$$

But I guess I cannot use complement.

So this would mean: $$A \cup B=X-(X-A-B)$$

Or, if $\cap$ has to be used: $$A \cup B=X-(X-(A-(A \cap B))-(B-(A \cap B))-(A \cap B))$$

calculatormathematical
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  • Given $A,B$ and the algebra $R$ they are subsets of, the answer is obvious, right? At least a few minutes of playing around should get you the complement of the set you want, and then you have it. – rschwieb Nov 11 '20 at 17:48
  • @rschwieb but i cannot use complement can i? – calculatormathematical Nov 11 '20 at 17:56
  • You can if you can mention $R$ (in @rschwieb's comment) but not if you cannot mention it. In fact $A-B=A\cap B$ if I understand you correctly, wherever the LHS is defined. Thus all you can ever get from subsets $A$ and $B$ is some subset of one or both. –  Nov 11 '20 at 18:03
  • @rschwieb What do you say about my idea? – calculatormathematical Nov 12 '20 at 08:49
  • @calculatormathematical I’ve not seen anyone try to use Boolean operations as partial operations on the entire class of sets. If that is your context, then no, I guess you don’t have complement . Are you sure you are not operating within the power set of some set? – rschwieb Nov 12 '20 at 11:31
  • @rschwieb I am not really sure. I only know the definition of this ,"monotone difference". I guess this could be right. – calculatormathematical Nov 12 '20 at 11:34
  • @calculatormathematical I've never heard the expression "monotone difference" so I don't know the context. It seems like it would be hard to make progress if you don't know either. – rschwieb Nov 12 '20 at 11:54

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